Title: CAT 2022 QA Slot 1: Analyzing an Indian Board Game Probability Question
Question (Hypothetical Example Based on CAT Pattern):
In a traditional Indian board game, players alternate turns to place a coin on a 4×4 grid. The first player to cover two adjacent rows or columns with their coins wins. If both players play optimally, what is the probability that Player A wins? Assume Player A starts first.
Solution:
Understanding the Game Mechanics:
The game is played on a 4×4 grid.
Players take turns placing coins on empty cells.
Player A wins by covering two adjacent rows or columns.
Optimal play implies neither player makes a move that allows the opponent to win immediately.
Key Observations:
The game ends when two adjacent rows or columns are fully occupied.
Total cells = 16. Each turn reduces the available cells by 1.
The maximum number of moves = 16 (if no player wins earlier).
Winning Condition Analysis:
To win, a player must occupy two adjacent rows (e.g., Rows 1 & 2) or columns (e.g., Columns 1 & 2).
Each row/column has 4 cells. Winning requires placing 8 coins in two adjacent rows/columns.
However, since players alternate, the game may end earlier if a player completes the required cells.
Simplified Strategy:
Assume the game progresses until all cells are filled (worst-case scenario).
The probability of Player A winning is the number of favorable move sequences divided by total possible sequences.
Combinatorial Calculation:
Total possible sequences = ( \binom{16}{8} ) (choosing 8 cells for Player A).
Favorable sequences = Number of ways Player A occupies two adjacent rows/columns in 8 moves.
There are 3 pairs of adjacent rows and 3 pairs of adjacent columns, totaling 6 winning configurations.
For each configuration, the number of ways = ( \binom{8}{8} \times \binom{8}{0} = 1 ) (since Player A must occupy all 8 cells).
Probability Expression:
[
P(\text{A wins}) = \frac{6 \times 1}{\binom{16}{8}} = \frac{6}{12,870} \approx 0.000466
]
Optimization Insight:
In reality, the game ends earlier when a player completes two adjacent rows/columns.
A dynamic programming approach would refine this probability by considering intermediate winning states.
However, for CAT’s time constraints, the combinatorial approximation suffices.
Final Answer:
[
\boxed{\frac{1}{2145}}
]
Note: The exact answer depends on precise game rules, but this solution aligns with CAT’s quantitative reasoning standards.
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